Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k^2 + 5k + 4}{-4k + 36} \div \dfrac{-9k - 9}{9k - 81} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{k^2 + 5k + 4}{-4k + 36} \times \dfrac{9k - 81}{-9k - 9} $ First factor the quadratic. $q = \dfrac{(k + 1)(k + 4)}{-4k + 36} \times \dfrac{9k - 81}{-9k - 9} $ Then factor out any other terms. $q = \dfrac{(k + 1)(k + 4)}{-4(k - 9)} \times \dfrac{9(k - 9)}{-9(k + 1)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k + 1)(k + 4) \times 9(k - 9) } { -4(k - 9) \times -9(k + 1) } $ $q = \dfrac{ 9(k + 1)(k + 4)(k - 9)}{ 36(k - 9)(k + 1)} $ Notice that $(k - 9)$ and $(k + 1)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ 9\cancel{(k + 1)}(k + 4)(k - 9)}{ 36(k - 9)\cancel{(k + 1)}} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $q = \dfrac{ 9\cancel{(k + 1)}(k + 4)\cancel{(k - 9)}}{ 36\cancel{(k - 9)}\cancel{(k + 1)}} $ We are dividing by $k - 9$ , so $k - 9 \neq 0$ Therefore, $k \neq 9$ $q = \dfrac{9(k + 4)}{36} $ $q = \dfrac{k + 4}{4} ; \space k \neq -1 ; \space k \neq 9 $